P1537 弹珠

用bitset位操作代替逐个询问

已存在,只需 存在,就有 存在

而题意只需要价值存在即可

利用鬼谷子的钱包的分治思想优化解法

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#include <cstdio>
#include <bitset>
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define LL long long
#define IN freopen("in.txt", "r", stdin)
#define OUT freopen("out.txt", "w", stdout)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define pr1(x) printf("Case %d: ", x)
#define pn1(x) printf("Case %d:\n", x)
#define pr2(x) printf("Case #%d: ", x)
#define pn2(x) printf("Case #%d:\n", x)
#define lowbit(x) (x & (-x))

typedef unsigned long long ull;

struct FastIO
{
static const int S = 5 << 20; //MB
int wpos;
char wbuf[S];
FastIO() : wpos(0) {}
inline int xchar()
{
static char buf[S];
static int len = 0, pos = 0;
if (pos == len)
pos = 0, len = fread(buf, 1, S, stdin);
if (pos == len)
return -1;
return buf[pos++];
}
inline int xuint()
{
int c = xchar(), x = 0;
while (~c && c <= 32)
c = xchar();
for (; '0' <= c && c <= '9'; c = xchar())
x = x * 10 + c - '0';
return x;
}
inline LL xull()
{
int c = xchar();
LL x = 0;
while (~c && c <= 32)
c = xchar();
for (; '0' <= c && c <= '9'; c = xchar())
x = x * 10 + c - '0';
return x;
}
inline int xint()
{
int s = 1, c = xchar(), x = 0;
while (c <= 32)
c = xchar();
if (c == '-')
s = -1, c = xchar();
for (; '0' <= c && c <= '9'; c = xchar())
x = x * 10 + c - '0';
return x * s;
}
inline void xstring(char *s)
{
int c = xchar();
while (c <= 32)
c = xchar();
for (; c > 32; c = xchar())
*s++ = c;
*s = 0;
}
inline void wchar(int x)
{
if (wpos == S)
fwrite(wbuf, 1, S, stdout), wpos = 0;
wbuf[wpos++] = x;
}
inline void wll(LL x)
{
if (x < 0)
wchar('-'), x = -x;

char s[30];
int n = 0;
while (x || !n)
s[n++] = '0' + x % 10, x /= 10;
while (n--)
wchar(s[n]);
}
inline void wstring(const char *s)
{
while (*s)
wchar(*s++);
}
~FastIO()
{
if (wpos)
fwrite(wbuf, 1, wpos, stdout), wpos = 0;
}
} io;

LL s, sum, a[7], cnt;
bitset<20000> dp;

int main()
{
//IN;
// OUT;
while (++cnt)
{
s = 0;
dp.reset();
dp.set(0);
f(i, 1, 6)
s += (a[i] = io.xint()) * i;
//io.wll(s);
if (s == 0)
break;
io.wstring("Collection #");
io.wll(cnt);
io.wchar(':');
io.wchar('\n');
if (s & 1)
io.wstring("Can't be divided.\n\n");
else
{
s >>= 1;
f(i, 1, 6)
{
for (int j = a[i]; j > 0; j >>= 1)
{
int k = ((j + 1) >> 1) * i;
dp |= (dp << k);
if (dp[s])
{
break;
}
}
if (dp[s])
break;
}
if (dp[s])
io.wstring("Can be divided.\n\n");
else
io.wstring("Can't be divided.\n\n");
}
}
}